Lakhmir Singh Manjit Kaur Chemistry 2020 2021 for Class 9 Science Chapter 3 - Atoms And Molecules

Answer:

The full form of IUPAC is International Union of Pure and Applied Chemistry.

Answer:

(a) Law of conservation of mass was proposed by Antoine Lavoisier.
(b) Law of constant proportions was proposed by Joseph Proust.

Answer:

(a) The law of chemical combination proposed by Lavoisier is well known as the law of conservation of mass.
(b) The law of chemical combination proposed by Proust is well known as the law of constant proportions.

Answer:

John Dalton proposed the atomic theory of matter.

Answer:

The postulate of Dalton's atomic theory, which is the result of the law of conservation of mass proposed by Lavoisier, states that, “Atoms can neither be created nor destroyed.”

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The postulate of Dalton's atomic theory which came from the law of constant proportions proposed by Proust, states that “The 'number' and 'kind' of atoms in a given compound is fixed.”

Answer:

Maharishi Kanada was the ancient Indian philosopher who suggested that  matter  comprises of very small particles. He named these particles ‘parmanu’.

Answer:

Two laws of chemical combination are:
  i.  Law of conservation of mass
  ii. Law of multiple proportions

Answer:

According to the given information, water has been taken from different sources and still decomposes into same amount of oxygen and hydrogen; therefore, this is an example of the law of constant proportions.

Answer:

If 100 grams of calcium carbonate is decomposed in any form (marble or chalk), 56 grams of calcium oxide and 44 grams of carbon dioxide are obtained. This reactions illustrates the law of conservation of mass.

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Atoms and molecules are the building blocks of matter.

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The size of an atom is indicated by its radius.

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The radius of an atom is usually expressed in nanometre (nm).

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1 nanometre = 1/10⁹ metre
or
1 nanometre = 10^–9 metre

Answer:

The symbol 'nm' in 0.073 nm (radius of an oxygen atom) represents the unit of atomic radius, i.e. nanometre.

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Atoms are the building blocks of nature. They are tiniest particles in nature and optical microscopes cannot magnify the image of such a tiny object, so that becomes visible to us. It is possible to watch an atom under an electron microscope.   

Answer:

False. The symbol for the element Cobalt is Co, as the second letter of the symbol of an element has to be in the lower case.

Answer:

The molecular mass of a substance is the relative mass of its one molecule as compared with a ${}_6{}^{12}\mathrm{C}$ atom. The mass of an atom of ${}_6{}^{12}\mathrm{C}$ is considered to be 12 a.m.u.

Answer:

The statement, "the molecular mass of oxygen is 32", means that a molecule of oxygen is 32 times heavier than 1/12^th of the mass of a ${}_6{}^{12}\mathrm{C}$ atom.

Answer:

(a) In water, the ratio of oxygen to hydrogen is 8:1 by mass.
(b) In a chemical reaction, the sum of masses of the reactants and their products remain unchanged. This is called the law of conservation of mass.

Answer:

(a) Carbon is used as a standard in the atomic mass scale.
(b) Carbon-12 atom is particularly used as a standard in the atomic mass scale.
(c) Carbon-12 atom has been given a mass of exactly 12 units (${}_6{}^{12}\mathrm{C}$ has six protons and six neutrons).

Answer:

Dalton's atomic theory states that atoms are indivisible; however, under special circumstances, atoms can be further divided into smaller particles called electrons, protons and neutrons. So, atoms are made of subatomic particles into which they can be broken or divided.

Answer:

Dalton’s atomic theory states that atoms are indivisible. This statement is not valid as, atoms can be further divided into electrons, protons and neutrons. This is one of the major drawbacks of Dalton’s atomic theory of matter.

Answer:

Yes, it is possible to see atoms these days, with the help of advanced technology. Electronic microscopes are extremely powerful and can magnify even the smallest particle. A Scanning Tunnelling Microscope (STM) can be used to produce highly magnified images of the surface of elements on a computer screen. Therefore, a highly sophisticated STM can be used to view atoms indirectly.

Answer:

The symbol of an element is either the first letter or the first and another letter of the English or Latin name of the element.
For example: Oxygen is represented by the symbol O. Similarly Iron, whose Latin name is Ferrum, is represented by the symbol Fe.

Answer:

(a) Two examples of symbols which have been derived from the “English names” of their respective elements: Sulphur – S and Aluminium – Al.

(b) Following are the two symbols which have been derived from the “Latin names” of their respective elements: Iron – Ferrum (Fe) and Copper (Cu)-Cuprum.

Answer:

The names of five familiar elements are:
• Oxygen:       O
• Iron:            Fe
• Gold:           Au
• Silver:         Ag
• Hydrogen:   H

Answer:

Following are the chemical symbols of the listed elements:
• Sodium:       Na
• Potassium:   K
• Iron:            Fe
• Copper:       Cu
• Mercury:     Hg
• Silver:          Ag

Answer:

Following are the chemical names for the listed symbols of elements:
• Hg: Mercury
• Pb: Lead
• Au: Gold
• Ag: Silver
• Sn: Tin

Answer:

Atomicity refers to the total number of atoms present in one molecule of an element.
For example, argon is a noble gas and exists in a free state. One molecule of argon comprises only one atom. Therefore, the atomicity of argon is one.
Similarly, a molecule of chlorine gas has two atoms and it exists as a Cl₂ molecule. Therefore, the atomicity of chlorine is two.

Answer:

The atomicity of the following:
(a) Oxygen:        2
(b) Ozone:          3
(c) Neon:            1
(d) Sulphur:       8
(e) Phosphorus: 4
(f) Sodium:         1

Answer:

A chemical formula is an expression that denotes the number and types of atoms present in a molecule of a substance. It is a representation of the composition of a molecule, in terms of the elements present, using their symbols.

For example: O₂ is the chemical formula of the element oxygen. Therefore, one molecule of oxygen comprises two atoms.

CO₂ is the chemical formula of the compound carbon dioxide. It consists of one atom of carbon and two atoms of oxygen.

Answer:

Answer:

N₂ is the chemical formula of nitrogen gas. It shows that two atoms of nitrogen form chemical bond to give one molecule of nitrogen gas.
2N, on the other hand, represents two atoms of nitrogen, which are not chemically bonded.

Answer:

(i) O stands for an atom of oxygen.
(ii) 2O refers to two atoms of oxygen which are not chemically bonded.
(iii) O₂ refers to a molecule of oxygen in which atoms are chemically bonded to give a molecule of oxygen gas.
(iv) 3O₂ refers to three molecules of oxygen.

Answer:

In H₂SO₄, H₂ represents two atoms of hydrogen; S represents an atom of sulphur; O₄ represents 4 atoms of oxygen where all these atoms are chemically bonded to each other to form a molecule of sulphuric acid i.e., H₂SO_4.

Answer:

(a) Naturally, oxygen  has diatomic molecules. Two oxygen atoms chemically combine with each other to give a molecule of O₂.

(b) Noble gases are present in the earth’s atmosphere. They exist as monoatomic gases.

Answer:

H₂ is the chemical formula of hydrogen gas. It denotes that two atoms of hydrogen chemically bond to give a molecule of hydrogen gas.
2H denotes the presence of two atoms of hydrogen, which are not chemically bonded.

Answer:

(i) N refers to an atom of nitrogen.
(ii) 2N represents two atoms of nitrogen which are not chemically bonded.
(iii) N₂ refers to a molecule of nitrogen gas. Two atoms of nitrogen chemically bond to give a molecule of nitrogen gas.
(iv)2N₂ refers to two molecules of nitrogen gas.

Answer:

The significance of the chemical formula of any substance is as follows:
• A chemical formula of a substance (element or compound) represents its name.
• A chemical formula provides important information on the chemical makeup of any substance (element or compound). It names all the elements present in the molecule.
• A chemical formula also provides important information on the number of atoms of each constituent element of the molecule of the substance.
• A chemical formula represents one mole of molecules of the substance (element or compound), which is 6.022 × 10²³ molecules of the substance.
• Since, a chemical formula represents one mole of molecules of a substance, it also represents the definite mass of that substance, which is expressed as gram molecular mass.

Answer:

The significance of the formula H₂O is:

• It represents one molecule of water.
• It provides important information on the constituents of water i.e. hydrogen and oxygen, through the chemical formula.
• It also provides information on the number of atoms of each element present in one molecule of water.
• It refers to the presence of one mole of molecules of water; it refers to 6.022 × 10²³ molecules of water.

Answer:

Given:
Atomic mass of C = 12 u
Atomic mass of H = 1 u
Atomic mass of O = 16 u
Therefore, molecular mass of C₆H₁₂O₆ can be calculated as follows:
(6 × 12) + (12 × 1) + (6 ×16) = 180
Hence, molecular mass of glucose is 180 u.

Answer:

(a) Given:
Atomic mass of H = 1 u
Therefore, molecular mass of H₂ can be calculated as follows:
1 × 2 = 2
Hence, molecular mass of H₂ is 2 u.

(b) Given:
Atomic mass of O = 16 u
Therefore, molecular mass of O₂ can be calculated as follows:
16 × 2 = 32
Hence, molecular mass of O₂ is 32 u.

(c) Given:
Atomic mass of Cl = 35.5 u
Therefore, molecular mass of Cl₂ can be calculated as follows:
35.5 × 2 = 71
Hence, molecular mass of Cl₂ is 71 u.

(d) Given:
Atomic mass of N = 14 u
Atomic mass of H = 1 u
Therefore, molecular mass of NH₃ can be calculated as follows:
(1 × 14) + (3 × 1) = 17
Hence, molecular mass of NH₃ is 17 u.

(e) Given:
Atomic mass of C = 12 u
Atomic mass of O = 16 u
Therefore, molecular mass of CO₂ can be calculated as follows:
(1 × 12) + (2 ×16) = 44
Hence, molecular mass of CO₂ is 44 u.

Answer:

(a) Given:
Atomic mass of C = 12 u
Atomic mass of H = 1 u
Therefore, molecular mass of CH₄ (methane) can be calculated as follows:
(1 × 12) + (4 × 1) = 16
Hence, molecular mass of methane is 16 u.

(b) Given:
Atomic mass of C = 12 u
Atomic mass of H = 1 u
Therefore, molecular mass of C₂H₆ (ethane) can be calculated as follows:
(2 × 12) + (6 × 1) = 30
Hence, molecular mass of ethane is 30 u.

(c) Given:
Atomic mass of C = 12 u
Atomic mass of H = 1 u
Therefore, molecular mass of C₂H₄ (ethylene) can be calculated as follows:
(2 × 12) + (4 × 1) = 28
Hence, molecular mass of ethylene is 28 u.

(d) Given:
Atomic mass of C = 12 u
Atomic mass of H = 1 u
Therefore, molecular mass of C₂H₂ (acetylene) can be calculated as follows:
(2 × 12) + (2 × 1) = 26
Hence, molecular mass of acetylene is 26 u.

Answer:

(a) Given:
Atomic mass of C = 12 u
Atomic mass of O = 16 u
Atomic mass of H = 1 u
Therefore, molecular mass of CH₃OH can be calculated as follows:
(1 ×12) + (4 × 1) + (1 × 16) = 32
Hence, molecular mass of methanol is 32 u.

(b) Given:
Atomic mass of C = 12 u
Atomic mass of O = 16 u
Atomic mass of H = 1 u
Therefore, molecular mass of C₂H₅OH can be calculated as follows:
(2 × 12) + (6 × 1) + (1 ×16) = 46
Hence, molecular mass of ethanol is 46 u.

Answer:

Given:
Atomic mass of C = 12 u
Atomic mass of O = 16 u
Atomic mass of H = 1 u
Therefore, molecular mass of CH₃COOH can be calculated as follows:
(2 × 12) + (4 × 1) + (2 ×16) = 60
Hence, molecular mass of acetic acid is 60 u.

Answer:

Given:
Atomic mass of H = 1 u
Atomic mass of N = 14 u
Atomic mass of O = 16 u
Therefore, molecular mass of HNO₃ can be calculated as follows:
(1 × 1) + (1 × 14) + (3 ×16) = 63
Hence, molecular mass of nitric acid is 63 u.

Answer:

Given:
Atomic mass of C = 12 u
Atomic mass of H = 1 u
Atomic mass of Cl = 35.5 u
Therefore, molecular mass of CHCl₃ can be calculated as follows:
(1 × 12) + (1 × 1) + (3 × 35.5) = 119.5
Hence, molecular mass of chloroform is 119.5 u.

Answer:

Given:
Atomic mass of H = 1 u
Atomic mass of Br = 80 u
Therefore, molecular mass of HBr can be calculated as follows:
(1 × 1) + (1 × 80) = 81
Hence, molecular mass of hydrogen bromide is 81 u.

Answer:

(a) Given:
Atomic mass of H = 1 u
Atomic mass of S = 32 u
Therefore, molecular mass of H₂S can be calculated as follows:
(2 × 1) + (1 × 32) = 34
Hence, molecular mass of hydrogen sulphide is 34 u.

(b) Given:
Atomic mass of C = 12 u
Atomic mass of S = 32 u
Therefore, molecular mass of CS₂ can be calculated as follows:
(1 × 12) + (2 × 32) = 76
Hence, molecular mass of carbon disulphide is 76 u.

Answer:

Law of conservation of mass states that the matter is neither created nor destroyed in a chemical reaction.
It was proposed by Antoine Lavoisier in 1774. It means that in a chemical reaction, the total mass of products (new substances that are formed in a chemical reaction) is equal to the total mass of reactants (substances which combine together in a chemical reaction).

Law of conservation of mass can be illustrated by the following chemical reaction where 100 grams of calcium carbonate decomposes to produce 56 grams of calcium oxide and 44 grams of carbon dioxide.


$\begin{array}{ccccc}\mathrm{Calcium} \mathrm{carbonate}& \to & \mathrm{Calcium} \mathrm{oxide}& +& \mathrm{Carbon} \mathrm{dioxide}\\ 100 \mathrm{gm}& & 56 \mathrm{gm}& & 44 \mathrm{gm}\end{array}$


Here,
Mass of reactant (calcium carbonate) = 100gm
Mass of products (calcium oxide & carbon dioxide) = 56gm + 44gm = 100gm
Since, the mass of reactant (100gm) is equal to the mass of the products (100gm), there is no change in mass in this chemical reaction. Therefore, this example illustrates the law of conservation of mass.

Answer:

Law of constant proportions states that a chemical compound always consists of the same elements combined together in the same proportion by mass. It was put forward by Joseph Proust in 1779
It means that a pure chemical compound is always made up of the same elements in the same mass percentage.

Law of constant proportions can be illustrated by the following example:

Carbon dioxide exists in nature in different forms and can be chemically produced in different ways. Yet, all the different samples of carbon dioxide will always contain the same elements i.e. carbon and oxygen in the same proportion by mass i.e. 3:8.

Answer:

(a) Postulates of Dalton's atomic theory of matter :

• Matter comprises very small particles called atoms.
• Atoms of different elements differ in mass, size and chemical properties.
• Atoms can neither be created nor destroyed.
• Chemical reactions involve the combination, separation or rearrangement of atoms to form molecules or compounds.
• Atoms cannot be divided.
• A compound has a fixed number and kind of atoms.
• Atoms of the same element can combine in more than one ratio to form more than one compound.
• All the atoms in an element are identical in every aspect i.e. they have same mass, size and chemical properties.
• Like elements, there are many kinds of atoms.
• Atoms of different elements join together to form molecules of the compound in a chemical combination.

(b) Dalton’s third postulate explains the law of conservation of mass. It states that atoms can neither be created nor destroyed.

(c)  Dalton’s postulate explains the law of constant proportions. It states that elements consist of atoms which have a fixed mass and also the number and kind of atoms in each element of a compound is fixed.

Answer:

(a) The symbol of an element is a short form used to represent individual elements or the atoms of the elements. Using symbols is advantageous, as using full names of the elements can be tedious. Symbols are commonly used to designate elements in a chemical reaction. Following are the significant points of using a symbol to designate an element:

• Symbol represents the name of the element. For example, the symbol O refers to the element oxygen.
Symbol represents one atom of the element. For example, the symbol O denotes one atom of the element oxygen.
• Using the symbol of an element refers to one mole of atoms of that element; therefore, the symbol of an element represents 6.022 × 10²³ atoms. For example, the symbol O represents one mole of oxygen  i.e. 6.022 × 10²³ atoms.
• Symbols of different elements represent a definite mass of those elements which is equal to gram atomic mass. For example, the symbol O refers to 1 gram atomic mass of oxygen.

(b) Significance of symbol H

• The symbol H refers to the element hydrogen.
• The symbol H refers to one atom of  hydrogen.
• The symbol H refers to one mole of hydrogen  i.e. 6.022 × 10²³ atoms.
• The symbol H refers to 1 gram atomic mass of hydrogen.

Answer:

(a) Atoms are building blocks of all matter. They are the smallest particles of elements that can take part in a chemical reaction.

Atoms usually exist in combined form, in two ways:
• in the form of molecules
• in the form of ions

(b) Molecule is the smallest particle of a substance (element or compound), which has the properties of that substance. It can exist in a free state.

Example: Oxygen atoms combine and exist in the form of O₃ molecules. Solid sulphur has eight sulphur atoms joined together, and exists in the form of S₈ molecule.

(c)

Answer:

(a) Atomic mass unit can be defined as one twelfth (1/12) of the mass of an unbound atom of carbon-12. The atomic masses of all the elements are determined by comparing the mass of their atom with that of the carbon-12 atom.

Atomic mass unit = 1/12 the mass of a carbon-12 atom

(b) The symbol of atomic mass unit is “u”.

(c) Atomic mass of an element refers to the relative mass of its atom compared to the mass of a carbon-12 atom. (Mass of carbon-12 atom is taken to be 12 u).

(d) By saying that the atomic mass of oxygen is 16, we mean that the atom of oxygen is 16 times heavier compared to 1/12^th the mass of a carbon-12 atom.

Answer:

(d)  3, 8, 4 and 1

Answer:

(b) Hg

The symbol of the metal used in making thermometers is Hg (Mercury).

Answer:

(a) Sodium

The English name of the element natrium is Sodium (Na).

Answer:

(d) John Dalton

The atomic theory of matter was proposed by John Dalton.

Answer:

(a) Helium
The atomicity of Helium (He) is 1.

Answer:

(c) kalium

The Latin name of the element potassium is kalium. Hence, the symbol for potassium is K.

Answer:

(c) Antoine Lavoisier

The law of conservation of mass was proposed by Antoine Lavoisier.

Answer:

(c) Phosphorus

Phosphorus has an atomicity of 4.

Answer:

(c) 2.5 gm and 1.1 gm

Molecular mass of CaCO₃: 40 + 12 + (16 × 3) = 100g
Molecular mass of CaO: 40 + 16 = 56g
Molecular mass of CO₂: 12 + (16 × 2) = 44g
Reaction for the decomposition of calcium carbonate:

$\begin{array}{ccccc}{\mathrm{CaCO}}_3& \to & \mathrm{CaO}& +& {\mathrm{CO}}_2\\ 100\mathrm{g}& & 56\mathrm{g}& & 44\mathrm{g}\\ \mathrm{X}& & 1.4\mathrm{g}& & \mathrm{Y}\end{array}$

56 g of CaO is obtained when 100g of CaCO₃ is burnt.
Therefore, to get 1.4g of CaO, the amount of CaCO₃ to be burnt will be
$\frac{100\times 1.4}{56}=2.5\mathrm{g}$
For CO₂,
100g of CaCO₃ yields 44g of CO₂
Therefore, 2.5g of CaCO₃ will yield;
$\frac{44\times 2.5}{100}=1.1 \mathrm{g}$

Answer:

(a) Proust

The law of constant proportion was proposed by Joseph Proust.

Answer:

(d) krypton and sulphur

Out of ozone, phosphorus, sulphur and krypton, krypton has the lowest atomicity and sulphur  the highest .

Answer:

(b) 10⁻⁷ cm

One nanometre is equal to 10⁻⁷ centimetre.

Answer:

(d) Berzelius

Berzelius proposed the first letter (or first letter and another letter) of the Latin or English name of any element as its symbol.

Answer:

(d) helium and neon

Atoms of helium and neon exist in a free state because they are inert gases and chemically inactive.

Answer:

(b) Neon

Neon’s molecular mass is equivalent to its atomic mass.

Answer:

(d) 8:1

In water, the ratio of oxygen to hydrogen is 8:1 by mass.

Answer:

(b) 1:16

In hydrogen peroxide, the ratio of hydrogen to oxygen is 1:16 by mass.

Answer:

(c)  He

He is the correct symbol of helium.

Answer:

Chemical reaction:
CuSO₄ + NaOH → Cu(OH)₂ + NaSO₄
15.94g      8.0g         9.75g        14.2g

Total mass of reactants = Total mass of products
 23.95g                                  23.95 g

The above reaction and data illustrate the law of conservation of mass which was proposed by Antoine Lavoisier.

The law states that the mass of elements remains constant over time i.e. mass can neither be created nor destroyed. In the above reaction, the mass of reactants (23.95 g) is exactly equal to the mass of their product (23.95 g). There is no change in the mass of the product after the reaction.

Answer:

Chemical reaction:

$\begin{array}{ccccc}2{\mathrm{KClO}}_3& \to & 2\mathrm{KCl}& +& 3{\mathrm{O}}_2\\ 24.5 \mathrm{g}& & 14.9 \mathrm{g}& & \end{array}$

So, mass of oxygen formed during the reaction:
Mass of reactants = 24.5 g
Mass of products = mass of the potassium chloride + mass of oxygen
24.5 = 14.9 + mass of oxygen
Mass of oxygen = 24.5 – 14.9 = 9.6 g
We have used the law of conservation of mass, proposed by Antoine Lavoisier, to solve this problem.

Answer:

To solve this problem, we need to work out the proportion of copper and oxygen in the two reactions.
1st chemical reaction:

$\begin{array}{ccccc}4\mathrm{Cu}& +& {\mathrm{O}}_2& \to & 2{\mathrm{Cu}}_2\mathrm{O}\\ 3.92 \mathrm{g}& & \mathrm{X} \mathrm{g}& & 4.90 \mathrm{g}\end{array}$

Therefore, mass of oxygen(X) = mass of copper oxide – mass of copper
Mass of oxygen = 4.9 – 3.92 = 0.98 g

In the above reaction, the ratio of mass of copper to the mass of oxygen is 3.92 : 0.98 = 4:1

2nd chemical reaction:
CuO + H₂  → Cu + H₂O
4.55g   Xg    3.64g  Yg

Molecular mass of copper oxide = 75.5g
Molecular mass of hydrogen = 2g
Molecular mass of copper = 63.5g
Molecular mass of water = 18g
Molecular mass of oxygen = 16g
Mass of hydrogen = 2/63.5 × 3.64 = 0.11g

Mass of water = 18/75.5 × 4.55 = 1.08g
Mass of oxygen in water = 16/18 × 1.08 = 0.96g

In the above reaction, the ratio of mass of copper to the mass of oxygen is 3.64 : 0.96 = 4:1
Ratio of copper to oxygen in both the reactions is 4:1. This illustrates and verifies the law of constant proportions.

Answer:

Magnesium reacts with oxygen in the ratio of 3:2 to form magnesium oxide.

The above relation states that magnesium and oxygen combine in a fixed ratio of 3:2 by mass.
Therefore, 3g of magnesium react with 2g of oxygen to form magnesium oxide.
So, 1g of magnesium requires = 2/3g of oxygen = 0.66g of oxygen
24g of magnesium require = 24 × 0.66g of oxygen = 16g
16g of oxygen would be required to react completely with 24g of magnesium.

Answer:

5g of calcium react with 2g of oxygen to form 7g of calcium oxide.

This problem can be solved using the law of constant proportions.
We know that calcium and oxygen react in a fixed ratio of 5:2, by mass, to produce 7g of calcium oxide. When 5g of calcium react with 20g of oxygen, the same amount of calcium oxide will be formed.

Therefore, 7g of calcium oxide will be formed when 5g of calcium will react with 20g of oxygen. Calcium and oxygen will react in a ratio of 5:2. The remaining oxygen will remain unreacted.
The law of constant proportions governs our answer.

Answer:

(a) (i) Liquid X is water (H₂O).
     (ii) Gas Y is oxygen.
    (iii) Gas Z is hydrogen.

(b) The ratio of element Z to element Y by mass = 22:178
Therefore, ratio of Z : Y by mass = 1:8

(c) The law of constant proportions is illustrated by this example.

(d) Liquid X has already been stated as water. Oceans and rivers are the important sources of water on earth.

(e) Y, oxygen, is essential for our survival. We breathe in oxygen during respiration.

Answer:

(a) (i) Solid P is calcium carbonate (CaCO₃).
     (ii) Gas Q is carbon dioxide (CO₂).
    (iii) Solid R is calcium oxide (CaO).

(b) When 50g of reactant P is heated, Q and R are produced. According to the law of conservation of mass, same amount of Q and R will be produced. Therefore, 50g of Q and R will be produced on heating 50g of P.

(c)  According to the law of conservation of mass, same amount of Q and R will be produced depending upon  the amount of P by mass.

(d) When we compare the mass of reactants to that of their product, it will be seen that the total mass of the reactants is equal to the total mass of their product.

(e) In the example, the law of conservation of mass proposed by Antoine Lavoisier is illustrated.

Answer:

The particles which have more or less electrons than the normal atoms are called ions.

Answer:

(a) Particles which have more electrons than the normal atoms are called anions.

(b) Particles which have fewer electrons than the normal atoms are called cations.

Answer:

Formula mass of a compound is the relative mass of its 'formula unit' compared to the mass of a carbon-12 atom (which has a mass of exactly 12 units).

Answer:

(a) Particles formed by the gain of electrons are called anions.
(b) Particles formed by the loss of electrons are called cations.

Answer:

(a) False. A sodium ion carries a positive charge because it has less number of electrons than a neutral atom.

(b) True. A chloride ion carries a negative charge because it has more electrons than a neutral atom.

Answer:

(a) Calcium oxide: CaO
(b) Magnesium hydroxide: Mg(OH)₂

Answer:

If Z has a valency of 3, then the formula of its oxide will be Z₂O₃.

Answer:

The particle that contains 10 electrons, 11 protons and 12 neutrons is a positively-charged sodium ion (Na⁺).

Answer:

The particle which has 18 electrons, 18 neutrons and 17 protons is a negatively-charged Chloride ion (Cl⁻).

Answer:

(a) Particle formed by the gain of electrons by its atom is called anion.
(b) Particle formed by the loss of electrons by its atom is called cation.
(c) Particle formed by losing or gaining electrons by its atom is called ion.
(d) A potassium ion carries a positive charge because it contains less number of electrons than protons.
(e) A sulphide ion carries a negative charge because it contains less number of protons than electrons.

Answer:

Water (H₂O) is made of Hydrogen and Oxygen.
Valency of Hydrogen is 1, and valency of Oxygen is 2.
Valency of one Hydrogen atom = 1
Valency of one Oxygen atom = 2
Therefore, to balance the valency of Oxygen atom, 2 atoms of Hydrogen atoms are required.
Hence, 2 atoms of Hydrogen will combine with 1 atom of Oxygen to form one molecule of water with the chemical formula H₂O.

Answer:

Valency of hydrogen = 1
Valency of nitrogen = 3
Ammonia consists of nitrogen and hydrogen.

$\begin{array}{ccc}\mathrm{Elements}:& \mathrm{N}& \mathrm{H}\\ \mathrm{Valencies}:& 3& 1\end{array}$

Formula for ammonia can be worked out by crossing the valencies of the constituent elements.

Therefore, formula of ammonia is NH₃.

Answer:

Valency of Sulphur = 4
Valency of Oxygen = 2

$\begin{array}{ccc}\mathrm{Elements}:& \mathrm{S}& \mathrm{O}\\ \mathrm{Valencies}:& 4& 2\end{array}$

Formula for sulphur dioxide can be worked out by crossing the valencies of the constituent elements.
Thus, formula of sulphur dioxide is S₂O₄. The formula has a common factor of 2; therefore, dividing the formula by the common factor gives us the simplest chemical formula: SO₂.

Answer:

Valency of carbon  =  4
Valency of sulphur = 2


$\begin{array}{ccc}\mathrm{Elements}:& \mathrm{C}& \mathrm{S}\\ \mathrm{Valencies}:& 4& 2\end{array}$

Formula for the compound containing carbon and sulphur can be worked out by crossing the valencies of the constituent elements.
Thus, formula of the compound (formed by the combination of carbon and sulphur) is C₂S₄. The formula has a common factor of 2; therefore, dividing the formula by the common factor gives us the simplest chemical formula, i.e., CS₂.
This compound is called carbon disulphide.

Answer:

Valency of element X = 4
Valency of element Y = 1

$\begin{array}{ccc}\mathrm{Elements}:& \mathrm{X}& \mathrm{Y}\\ \mathrm{Valencies}:& 4& 1\end{array}$

Formula for the compound formed between X and Y can be worked out by crossing  the valencies of the constituent elements.

Therefore, formula of the compound is XY₄.

Answer:

Case i

Valency of B = 4
Valency of O = 2

$\begin{array}{ccc}\mathrm{Elements}:& \mathrm{B}& \mathrm{O}\\ \mathrm{Valencies}:& 4& 2\end{array}$

Formula for the oxide formed by the element B (valency 4) can be worked out by crossing over the valencies of the constituent elements.
Therefore, formula of the oxide of element B is B₂O₄. The formula has a common factor of 2; hence, dividing the formula by the common factor gives us the simplest chemical formula: BO₂.

Case ii

Valency of element B = 6
Valency of O = 2

$\begin{array}{ccc}\mathrm{Elements}:& \mathrm{B}& \mathrm{O}\\ \mathrm{Valencies}:& 6& 2\end{array}$

Formula for the oxide formed by the element B (valency 6) can be worked out by crossing over the valencies of the constituent elements.
Therefore, formula of the oxide of element B is B₂O₆. The formula has a common factor of 2; therefore, dividing the formula by the common factor gives us the simplest chemical formula: BO₃.

Answer:

Valency of element X = 3
Valency of element Y = 2

$\begin{array}{ccc}\mathrm{Elements}:& \mathrm{X}& \mathrm{Y}\\ \mathrm{Valencies}:& 3& 2\end{array}$

Formula for the compound formed between X and Y can be worked out by crossing the valencies of the constituent elements.
Therefore, formula of the compound is X₂Y₃.

Answer:

Valency of Magnesium = 2
Valency of hydrogen carbonate ion = −1

$\begin{array}{ccc}\mathrm{Elements}/\mathrm{Ions}:& \mathrm{Mg}& {\mathrm{HCO}}_3^{-}\\ \mathrm{Valencies}:& +2& -1\end{array}$

Formula for Magnesium hydrogen carbonate can be worked out by


Therefore, formula for Magnesium hydrogen carbonate is: Mg(HCO₃)₂.

Answer:

(a)

Valency of X = 2
Valency of Br = 1

Formula for bromide of element X can be worked out by crossing the valencies of the constituent elements.
Therefore, formula for bromide of X is: XBr₂.

(b)

Valency of X = 2
Valency of Oxygen = 2

Formula for oxide of element X can be worked out by crossing the valencies of the constituent elements.
Therefore, formula for oxide of X is XO.

Answer:

(a) Valency of Sodium = 1
Valency of Oxygen = 2



Formula for sodium oxide can be worked out by crossing the valencies of the constituent elements.
Therefore, formula for sodium oxide is: Na₂O.

(b) Valency of Calcium = 2
Valency of Carbonate ion = −2



Formula for calcium carbonate can be worked out by crossing  the valencies of the constituent elements/ions. +2 positive charge of calcium neutralizes the −2 negative charge of carbonate ion.
Therefore, formula for calcium carbonate is: CaCO₃.

Answer:

(i) Formula mass of Na₂O:

Formula mass of Na₂O = Mass of two Na atoms + Mass of one O atom

Answer:

(a) Copper sulphate (Copper ion – Cu²⁺ and Sulphate ion – SO₄²⁻)
(b) Ammonium sulphate (Ammonium ion −NH₄⁺ and Sulphate ion −SO₄²⁻)
(c) Sodium oxide (Sodium ion −Na⁺ and Oxygen ion −O²⁻)
(d) Sodium carbonate (Sodium ion −Na⁺ and Carbonate ion −CO₃²⁻)
(e) Calcium chloride (Calcium ion −Ca²⁺ and Chloride ion −Cl⁻)

Answer:

(a)$\begin{array}{ccccc}{\mathrm{CH}}_3\mathrm{COONa}:& \mathrm{Cation}:& {\mathrm{Na}}^{+}& \mathrm{Anion}:& {\mathrm{CH}}_3{\mathrm{COO}}^{-}\end{array}$
(b) $\begin{array}{ccccc}\mathrm{NaCl}:& \mathrm{Cation}:& {\mathrm{Na}}^{+}& \mathrm{Anion}:& {\mathrm{Cl}}^{-}\end{array}$
(c) H₂ is a neutral molecule; therefore, it will not have a cation or an anion.
(d)$\begin{array}{ccccc}{\mathrm{NH}}_4{\mathrm{NO}}_3:& \mathrm{Cation}:& {\mathrm{NH}}_4^{+}& \mathrm{Anion}:& {\mathrm{NO}}_3^{-}\end{array}$

Answer:

(a) Compound formed with calcium and fluorine will have the chemical formula CaF₂. (valency of Ca is 2 and that of F is 1)
(b) Compound formed with hydrogen and sulphur will have the chemical formula H₂S. (valency of S is 2 and that of H is 1)
(c) Compound formed with nitrogen and hydrogen will have the chemical formula NH₃. (valency of N is 3 and that of H is 1)
(d) Compound formed with carbon and chlorine will have the chemical formula CCl₄. (valency of C is 4 and that of Cl is 1)
(e) Compound formed with sodium and oxygen will have the chemical formula Na₂O. (valency of Na is 1 and that of O is 2)
(f) Compound formed with carbon and oxygen will have the chemical formula CO₂. (valency of C is 4 and that of O is 2, divided by the common factor of 2)

Answer:

Ionic Compounds: Ionic compounds that consist of oppositely-charged ions held together by strong electrostatic forces. In an ionic compound, positively-charged ions, cations, and negatively-charged ions, anions, are chemically bonded by ionic bonds.
Examples of ionic compounds: NaCl – It comprises positively-charged sodium ions chemically bonded to the negatively-charged chloride ions.
Similarly, CuSO₄  comprises positively-charged copper ions chemically bonded to the negatively-charged sulphate ions.
Molecular compounds: Molecular compounds consist of molecules of two or more elements that are chemically bonded. Molecular compounds are made up of electrically-neutral molecules and not electrically-charged ions.
Examples of molecular compounds: SO₂ – It consists of electrically-neutral sulphur chemically bonded to two atoms of oxygen.
Similarly, NH₃ consists of electrically-neutral nitrogen atom chemically bonded to three atoms of hydrogen.

Answer:

(a) An ion is an electrically-charged atom or molecule in which the total number of electrons is not equal to the total number of protons.
An ion is formed when a neutral atom loses or gains electrons. This loss or gain of electrons imparts a charge (positive or negative) to the neutral atom making it an ion.
There are two different types of ions:
If a neutral atom loses an electron, an overall positive charge is imparted to the atom and it becomes a positively-charged ion. A positively charged ion is called a cation. In a cation, the total number of electrons is less than the total number of protons as compared to the neutral atom. For example, sodium readily loses an electron to become a positively-charged sodium ion (Na⁺).
If a neutral atom gains an electron, an overall negative charge is imparted to the atom and it becomes a negatively-charged ion. A negatively-charged ion is called an anion. It has  more no of electrons than the  number of protons compared to the neutral atom. For example, chlorine readily gains an electron to become a negatively-charged chloride ion (Cl^-).

(b) Formula of the listed compounds are as follows:
Sodium phosphate –      Na₃PO₄
Ammonium sulphate –   (NH₄)₂SO₄
Calcium hydroxide –     Ca(OH)₂
Lead bromide –            PbBr₂

Answer:

Answer:

Formation of Sodium ion:
A neutral sodium atom consists of 11 protons and 11 electrons. The protons and electrons are equal in number, thus keeping the overall charge of the electrically neutral sodium atom zero.
A sodium ion is formed when the neutral atom loses an electron.

Answer:

(a) Example of a simple ion: Iodine atom forms an iodide ion by accepting an electron, I^-. Similarly, magnesium loses two electrons to form a magnesium ion, Mg²⁺.
Example of a compound ion: Carbonate ion, CO₃^2-, is made up of two types of atoms i.e. carbon and oxygen (chemically bonded). Similarly, ammonium ion, NH₄⁺, is made up of two types of atoms i.e. nitrogen and hydrogen (chemically bonded).

(b) Formula for the listed compounds are as follows:
(i) Crossing the valencies, chloride for the element Y will have the chemical formula YCl₄.


(ii) Crossing over the valencies, oxide for the element Y will have the chemical formula YO₂.


(iii) Crossing the valencies, sulphate for the element Y will have the chemical formula Y(SO₄)₂.


(iv) Crossing the valencies, carbonate for the element Y will have the chemical formula Y(CO₃)₂.


(v)Crossing  the valencies, nitrate for the element Y will have the chemical formula Y(NO₃)₄.

Answer:

(a) Formula unit of an ionic compound refers to the empirical formula of an ionic or covalent molecule. It is the lowest whole number ratio of ions that make up the electrically neutral ionic compound. Formula unit represents the smallest unit of a compound.
The formula unit of sodium chloride is NaCl.
The formula unit of magnesium chloride is MgCl₂.

(b) (i) Formula mass of calcium chloride = mass of one Ca atom + mass of two Cl atoms = 40 + 2 × 35.5 = 111
Thus, the formula mass of calcium chloride is 111 u.
(ii) Formula mass of sodium carbonate = mass of two Na atoms + mass of one C atom + mass    of three O atoms = 2 × 23 + 12 + 3 × 16 = 106
Thus, the formula unit of sodium carbonate is 106 u.

Answer:

(d) 10
If the atomic number of X is 13, the number of electrons in its X³⁺ ion would be 13 – 3 = 10.

Answer:

(b) Na₃N
The correct chemical formula among the listed chemicals is of Na₃N.

Answer:

(a) 7
If the number of electrons in an ion Z^3– is 10, the atomic number of element Z will be 7 because the ion is formed by gaining three electrons.

Answer:

(a) more electrons than the normal atom

The anion of an element has more electrons than the normal atom because it is formed by gaining electrons.

Answer:

(b) anion
X, most likely, is an anion because the number of electrons is more than the number of protons.

Answer:

(d) Sulphur
Sulphur exhibits the valencies of 2, 4 and 6.

Answer:

(b) 18
Element with an atomic number 16 will have 18 electrons in its E^2- ion because anions are formed by gaining electrons.

Answer:

(d) number of electrons less than the neutral atom 

The cation of an element has less electrons than a neutral atom.

Answer:

(d) nitrogen and iron
Elements X and Y, most likely, are nitrogen and iron.

Answer:

(b) 12
The atomic number of element Y would be 12.

Answer:

(d) cation
Particle P must be a cation.

Answer:

(a) Chlorine and calcium
Chlorine and calcium can form an ionic compound CaCl₂.

Answer:

(b) two different non-metals
Molecular compounds are formed by the combination of two different non-metals.

Answer:

(a) 1 and 3

Valencies of element X and Y will be 1 and 3, respectively.

Answer:

(c) XN
Formula of nitride of the element X will be XN.

Answer:

(a) Element A forms an oxide A₂O₅. Crossing the valencies, we can see that the valency of O (oxide) is −2 and that of element A is 5.

(b) Formula of chloride of A:
$\begin{array}{ccc}\mathrm{Element}/\mathrm{Ion}& \mathrm{A}& {\mathrm{Cl}}^{-}\\ \mathrm{Valency}& 5& -1\end{array}$
Formula of the chloride of element A can be worked out by crossing over the valencies. Thus, the formula is ACl₅.

Answer:

H₂X:
$\begin{array}{ccc}\mathrm{Element}/\mathrm{Ion}& \mathrm{H}& \mathrm{X}\\ \mathrm{Valency}& 1& 2\end{array}$
To stabilize the compound with hydrogen, two atoms of X are required. Therefore, the valency of X in this compound will be two.

CX₂:
$\begin{array}{ccc}\mathrm{Element}/\mathrm{Ion}& \mathrm{C}& \mathrm{X}\\ \mathrm{Valency}& 4& 2\end{array}$
To stabilize the compound with carbon, two atoms of X are required. Thus, the valency of X in this compound is two.

XO₂:
$\begin{array}{ccc}\mathrm{Element}/\mathrm{Ion}& \mathrm{X}& \mathrm{O}\\ \mathrm{Valency}& 4& -2\end{array}$
To stabilize this compound with oxygen, four atoms of X are required. Thus, the valency of X in this compound is four.

XO₃:
$\begin{array}{ccc}\mathrm{Element}/\mathrm{Ion}& \mathrm{X}& \mathrm{O}\\ \mathrm{Valency}& 6& -2\end{array}$
To stabilize this compound with oxygen, six atoms of X are required. Thus, the valency of X in this compound  is six.

Answer:

Valency of X in the aluminium salt Al₂X₃ can be worked out as follows (we know that valency of Al is 3):

$\begin{array}{ccc}\mathrm{Element}& \mathrm{Al}& \mathrm{X}\\ \mathrm{Valency}& 3& 2\end{array}$

Thus, it is clear that the valency of X is two.

Formula for the magnesium salt of X can be worked out as follows:
$\begin{array}{ccc}\mathrm{Element}& \mathrm{Mg}& \mathrm{X}\\ \mathrm{Valency}& 2& 2\end{array}$
Crossing the valencies gives us the formula of the magnesium salt of X i.e. MgX.

Answer:

Let us first work out the valency of the metal M (we know that carbonate ion has a valency of −2):
$\begin{array}{ccc}\mathrm{Element}/\mathrm{Ion}& \mathrm{M}& {\mathrm{CO}}_3^{2-}\\ \mathrm{Valency}& 1& 2\end{array}$
It is clear that M has a valency of 1.

(a) Formula of iodide of M:
$\begin{array}{ccc}\mathrm{Element}& \mathrm{M}& \mathrm{I}\\ \mathrm{Valency}& 1& 1\end{array}$
Crossing  the valencies, formula of the iodide of M is MI.

(b) Formula of nitride of M:
$\begin{array}{ccc}\mathrm{Element}/\mathrm{Ion}& \mathrm{M}& {\mathrm{N}}^{3-}\\ \mathrm{Valency}& 1& -3\end{array}$
Crossing the valencies, formula of the nitride of M is M₃N.

(c) Formula of phosphate of M:
$\begin{array}{ccc}\mathrm{Element}/\mathrm{Ion}& \mathrm{M}& {\mathrm{PO}}_4^{3-}\\ \mathrm{Valency}& 1& -3\end{array}$
Crossing the valencies, formula of the phosphate of M is M₃PO₄.

Answer:

(a) Element X consists of 17 protons and electrons and 18 neutrons. Element X will readily accept an electron to form a negatively charged anion.
Symbol of the negatively charged X is X⁻.

(b) In the ion formed by X:
  (i)  Number of protons = 17
  (ii) Number of electrons = 18
  (iii) Number of neutrons = 18

(c) Element Y consists of 11 protons and electrons and 12 neutrons. Element Y will readily lose an electron to form a positively charged cation.
Symbol of the positively charged Y is Y⁺.

(d) In the ion formed by Y:
(i) Number of protons = 11
(ii) Number of electrons = 10
(iii) Number of neutrons = 12

(e) (i) Atomic mass of X = number of protons of X + number of neutrons of X = 17 + 18 = 35 u

(ii) Atomic mass of Y = number of protons of Y + number of neutrons of Y = 11 + 12 = 23 u

(f) Atomic number of X is 17; therefore, element X is chlorine.
Atomic number of Y is 11; therefore, element Y is sodium.

Answer:

A group of 6.022 × 10²³ particles is known as one mole.

Answer:

The name given to the amount of substance containing 6.022 × 10²³ particles is 1 mole.

Answer:

The numerical value of Avogadro’s number is 6.022 × 10²³.

Answer:

6.022 × 10²³ (avogadro's number) atoms are present in one gram atomic mass of a substance.

Answer:

6.022 × 10²³ molecules are present in one gram molecular mass of a substance.

Answer:

The name given to the number 6.022 × 10²³ is Avogadro’s number.

Answer:

Gram molecular mass of O₂ = 32 g
mass of oxygen gas given = 12 g
Therefore, 32 g of O₂ = 1 mole of O₂
So, 12 g of O₂ = $\frac{12}{32}$ moles of O₂
Hence, 12 g of oxygen contains 0.375 moles of oxygen.

Answer:

Given  mass of water = 3.6g
Gram molecular mass of water = 18g
Therefore, 18g of water = 1 mole of water
So, 3.6g of water = $\frac{3.6}{18}$ moles of water
Hence, 3.6g of water contain 0.2 moles of water.

Answer:

Given mass of oxygen =  0.2 moles
Gram atomic mass of oxygen atom = 16g
Therefore, 16g of oxygen atoms = 1 mole of oxygen atoms
So, 0.2 moles of oxygen atoms = 16 × 0.2g of oxygen atoms
Hence, 0.2 moles of oxygen atoms weigh 3.2g.

Answer:

Given moles on N atom = 2 moles
Gram atomic mass of nitrogen atoms = 14g
Therefore, 14g of nitrogen atoms = 1 mole of nitrogen atoms
So, 2 moles of nitrogen atoms = 14 × 2g of nitrogen atoms
Hence, 2 moles of nitrogen atoms weigh 28g.

Answer:

(a) 1 mole contains 6.022 × 10²³ atoms, molecules or ions of a substance.
(b) A mole represents the Avogadro number of particles of a substance.
(c) 60g of carbon element are 5 moles of carbon atoms.
(d) 0.5 moles of calcium element have a mass of 20g.
(e) 64g of oxygen contain 4 moles of oxygen atoms.

Answer:

(a) 6.022 × 10²³ atoms are present in 12g of carbon-12 element.
(b) The name given to this number is Avogadro’s number.
(c) A substance containing 6.022 × 10²³ number of atoms is called 1 mole of that substance.

Answer:

12.044 × 10²⁵ molecules of oxygen
Mass of 6.022 × 10²³ molecules of oxygen = 32g
Mass of one molecule of oxygen = 32 / 6.022 × 10²³ = 5.3 × 10⁻²³g
Mass of 12.044 × 10²⁵ molecules of oxygen = 5.3 × 10⁻²³ × 12.044 × 10²⁵ g = 6400g or 6.4kg
Hence, the mass of 12.044 × 10²⁵ molecules of oxygen is 6.4kg.

Answer:

1.5 moles of ammonia
Number of molecules in 1 mole of ammonia = 6.022 × 10²³
Number of molecules in 1.5 moles of ammonia = 1.5 × 6.022 × 10²³
Hence, number of molecules in 1.5 moles of ammonia is 9.033 × 10²³.

Answer:

Mass of calcium carbonate = 10g
Molecular mass of calcium carbonate = (1 × 40) + (1 × 12) + (3 × 16) = 100g
Number of moles in 100g of calcium carbonate = 1 mole
Therefore, number of moles in 10g of calcium carbonate is 0.1.

Answer:

1.20 × 10²² molecules of oxygen

One mole of oxygen contains 6.022 × 10²³ molecules.
1.20 × 10²² molecules of oxygen contain = 1.20 × 10²² / 6.022 × 10²³ = 0.01099 moles.
Hence, 1.20 × 10²² molecules of oxygen contain 0.0199 moles of oxygen.

Answer:

Mass of one mole of nitrogen = 28g
We know that, 1 mole of nitrogen contains 6.022 × 10²³ molecules.
Mass of 6.022 × 10²³ nitrogen molecules = 28g
Mass of one molecule of nitrogen = 28 / 6.022 × 10²³ g = 4.64 × 10^-23 grams.
Hence, one molecule of nitrogen weighs 4.64 × 10^-23 grams.

Answer:

Mass of sodium = 34.5g
We know that 1 mole of sodium weighs 23g.
Therefore, 34.5g of sodium  = 34.5 / 23 = 1.5 moles
Hence, 34.5g of sodium contain 1.5 moles of sodium.

Answer:

Mass of zinc = 10g
gram atomic mass of zinc = 65g
We know that one mole of zinc contains 6.022 × 10²³ atoms of zinc.
65g of zinc contain = 6.022 × 10²³ atoms
Therefore, 10g of zinc contain = 10 × 6.022 × 10²³/65 = 9.264 × 10²² atoms

Hence, 10 grams of zinc contain 9.264 × 10²² atoms.

Answer:

Number of carbon atoms = 3.011 × 10²⁴
We know that 12g (atomic mass of carbon) of carbon contains 6.022 × 10²³ carbon atoms.
Mass of 1 atom of carbon = 12/6.022 × 10²³ = 1.99 × 10^-23
Hence, mass of 3.011 × 10²⁴ carbon atoms = 3.011 × 10²⁴ ×1.99 × 10^-23 = 59.9g or 60g (approximately)

Answer:

Given:
Mass of 1 mole of oxygen = 16g
We know that 1 mole of oxygen contains 6.022 × 10²³ oxygen atoms.
Mass of 6.022 × 10²³ oxygen atoms = 16g
Mass of 1 oxygen atom = 16/6.022 × 10²³
Hence, 1 atom of oxygen molecule weighs 2.65 × 10^-23 grams.

Answer:

0.25 moles of hydrogen
Number of hydrogen atoms in 1 mole = 6.022 × 10²³
Therefore, number of atoms in 0.25 moles = 6.022 × 10²³ × 0.25 = 1.50 × 10²³ atoms
Hence, number of hydrogen atoms in 0.25 moles is 1.50 × 10²³

Answer:

12.044 × 10²⁵ atoms of phosphorus
6.022 × 10²³ atoms of phosphorus = 1 mole
12.044 × 10²⁵ atoms of phosphorus = 12.044 × 10²⁵/6.022 × 10²³ = 199.9 moles or 200 moles
Hence, 12.044 × 10²⁵ atoms of phosphorus have 200 moles.

Answer:

Mass of chloroform = 0.0239g
Molar mass of chloroform = (1 × 12) + (1 × 1) + (3 × 35.5) = 119.5g
Number of molecules in 119.5g of chloroform = 6.022 × 10²³
Number of molecules in 1g of chloroform = 6.022 × 10²³/119.5
Number of molecules in 0.0239g of chloroform = 0.0239 × 6.022 × 10²³/119.5     
Hence, a drop of chloroform weighing 0.0239g has 1.2 × 10²⁰ molecules.

Answer:

Number of moles of sodium carbonate = 5 moles
Molar mass of sodium carbonate = (2 × 23) + (1 × 12) + (3 × 16) = 106g
We know that 1 mole of sodium carbonate weighs 106g.
Mass of 5 moles of sodium carbonate = 5 × 106 = 530g
Hence, 5 moles of sodium carbonate weigh 530 grams.

Answer:

Mass of oxygen = 4g
We know that 32g of oxygen contain 6.022 × 10²³ molecules.
Number of molecules in 32g of oxygen = 6.022 × 10²³
Number of molecules in 4g of oxygen = 4 × 6.022 × 10²³/32
Hence, 4g of oxygen have 7.52 × 10²² molecules.

Answer:

Mass of glucose = 100g
Molar mass of glucose    = (6 × 12) + (12 × 1) + (6 × 16) = 180g
1 mole of glucose = 180g
Or, 180g of glucose = 1 mole of glucose
Therefore, 100g of glucose = 100 / 180 moles = 0.55 moles
Hence, 100g of glucose contain 0.55 moles.

Answer:

Moles of hydrogen sulphide = 0.17
Molar mass of hydrogen sulphide = (2 × 1) + (1 × 32)g = 34g                    
1 mole of hydrogen sulphide weighs = 34g
Therefore, mass of 0.17 moles of hydrogen sulphide = 34 × 0.17 = 5.78g
Hence, 0.17 moles of hydrogen sulphide weigh 5.78 grams.

Answer:

The mass of 5 moles of carbon dioxide and 5 moles of water is not the same. This has been proved below:
Mass of 5 moles of carbon dioxide:
Molar mass of carbon dioxide = (1 × 12) + (2 × 16) = 44g
We know that 44g of carbon dioxide represent 1 mole.
Therefore, 5 moles of carbon dioxide weigh = 5 × 44 = 220g
Mass of 5 moles of water:
Molar mass of water = (2 × 1) + (1 × 16) = 18g
We know that 18g of water represent 1 mole.
Therefore, 5 moles of water weigh = 5 × 18 = 90g
Hence, it is proved from the above calculation that the mass of 5 moles of carbon dioxide and water is not the same. 5 moles of carbon dioxide weigh 130g more than 5 moles of water.

Answer:

Mass of calcium = 240g
Molar mass of calcium = 40g
Number of moles in 40g of calcium = 1 mole
Therefore, number of moles in 240g of calcium = 240 / 40 = 6 moles
Mass of magnesium = 240g
Molar mass of magnesium = 24g
Number of moles in 24g of magnesium = 1 mole
Therefore, number of moles in 240g of magnesium = 240 / 24 = 10 moles
Therefore, mole ratio of 240g of calcium to 240g of magnesium is 3:5.

Answer:

(a) Mole can be defined as, the amount of a pure substance, containing the same number of atoms, molecules or ions as present in 12g of carbon. This refers to the value of 6.022 × 10²³ atoms, molecules or ions of a substance.
A mole represents Avogadro number of particles of a substance.

(b)
1.5 moles of sodium sulphite
Molar mass of sodium sulphite = (2 × 23) + (1 × 32) + (3 × 16) = 126g    
Therefore, 1 mole of sodium sulphite weighs 126g.
Now, 1 mole of sodium sulphite contains 2 moles of sodium.
So, mass of sodium in 1 mole of sodium sulphite = 2 × 23 = 46g
Therefore, mass of sodium in 1.5 moles of sodium sulphite is = 1.5 × 46 = 69g

1 mole of sodium sulphite contains 1 mole of sulphur.
So, mass of sulphur in 1 mole of sodium sulphite = 1 × 32 = 32g
Therefore, mass of sulphur in 1.5 moles of sodium sulphite is = 1.5 × 32 = 48g

1 mole of sodium sulphite contains 3 moles of oxygen.
So, mass of oxygen in 1 mole of sodium sulphite = 3 × 16 = 48g
Therefore, mass of oxygen in 1.5 moles of sodium sulphite is = 1.5 × 48 = 72g
Hence, the mass of sodium, sulphur and oxygen in sodium sulphite is 69g, 48g and 72g, respectively.

Answer:

(a) A mole can be defined as the amount of a pure substance containing the same number of atoms, molecules or ions as present in exactly 12g of carbon. This refers to the value of 6.022 × 10²³ atoms, molecules or ions of a substance.
Therefore, a mole of carbon atoms refers to a group of 6.022 × 10²³ carbon atoms.

(b)
Mass of aluminium = 50g
We know that 27g of aluminium contain 6.022 × 10²³ aluminium atoms
Number of aluminium atoms in 27g  aluminium = 6.022 × 10²³
Number of aluminium atoms in 50g aluminium = 50 × 6.022 × 10²³/ 27
Hence, 50g aluminium has11.15 × 10²³ atoms.

Mass of iron = 50g
We know that 56g  iron contain 6.022 × 10²³ atoms
Number of atoms in 56g iron = 6.022 × 10²³
Number of atoms in 50g  iron = 50 × 6.022 × 10²³ / 56
Hence, 50g iron comprises 5.376 × 10²³ atoms.
It is clear that more number of atoms is present in 50g  aluminium than 50g  iron.

Answer:

(a) Gram atomic mass refers to the mass of an element , in one mole of atoms of that substance. It is equal to the gram atomic weight of that element.
Gram atomic mass of oxygen is 16 grams.

(b) Number of moles of oxygen in the listed compounds are as follows:
• Al₂O₃:  3 moles of oxygen are present in 1 mole of Al₂O₃.
• CO₂:  2 moles of oxygen are present in 1 mole of CO₂.
• Cl₂O₇: 7 moles of oxygen are present in 1 mole of Cl₂O₇.
• H₂SO₄:  4 moles of oxygen are present in 1 mole of H₂SO₄.
• Al₂(SO₄)₃: 12 moles of oxygen are present in 1 mole of Al₂(SO₄)₃ (4 × 3 moles).

Answer:

(a) Gram molecular mass refers to the mass of a molecules in one mole of molecules of that substance. It is equal to the gram molecular weight of that element.
Oxygen exists as a diatomic molecule (O₂); therefore, gram molecular mass of oxygen is 32 grams.

(b)
Mass of sulphur = 100g (sulphur exists as S8 molecules)
molecular mass of sulphur = (8 × 32) = 256g
Number of moles in 256g of sulphur  = 1 mole
Therefore, number of moles in 100g of sulphur = 100/256 = 0.39 moles
Hence, 100g of sulphur contains 0.39 moles of sulphur.

Answer:

(a) Molar mass of a substance (molecular compound) is the mass in grams, numerically equivalent to the sum of atomic masses of individual atoms in the molecular formula of that substance.
Molar mass of a substance is usually expressed as grams/mole.

(b)

(i) Molar mass of ozone (O₃)
Molar mass of ozone = 16 + 16 +16 = 48 g/mol

(ii) Molar mass of acetic acid (CH₃COOH)
Molar mass of acetic acid = (2 × 12) + (4 × 1) + (2 × 16) = 60 g/mol

Answer:

(b) 12 g of magnesium and 6 g of carbon
      12 g of magnesium and 6 g of carbon will have a mole ratio of 1 : 1.

Answer:

(d) (ii) and (iv)
20 moles of water and 1.2044×10²⁵ molecules of water correctly represent 360 gms of water.

Answer:

(b) 2x
The number of oxygen atoms in 32 g of oxygen will be 2x.

Answer:

(d) 12g of magnesium and 6g of carbon
The student will have to weigh 12 g of Magnesium and 6 g of Carbon.

Answer:

(c) 1:1
The ratio of moles of atoms in 12 g of Magnesium and 16 g of Sulphur will be 1:1.

Answer:

(c) The number of atoms in 12g of magnesium will be x/2.

Answer:

(d) 18g of CH₄ has  maximum number of atoms.

Answer:

Given:
Mass of sulphur dioxide = 1g
We know that equal moles of all substances contain equal number of molecules. (1 mole of all substances contains 6.022 × 10²³ molecules).
Therefore, the ratio of molecules in sulphur dioxide and oxygen will be same as the ratio of their moles.
Number of moles in 1g of sulphur dioxide:
1 mole of SO₂ = (1 × 32) + (2 × 16) = 64g
64g of SO₂ = 1 mole of SO₂
Therefore, 1g of SO₂ = 1/64 moles of SO₂
Now, 1/64 moles of SO₂ contain x molecules (given). Since equal number of molecules are present in equal moles of all substances, 1/64 moles of oxygen will also contain x number of molecules.
Number of moles in 1g of oxygen:
1 mole of O₂ = 2 × 16 = 32g
32g of O2 = 1 mole of O₂
Therefore, 1g of O₂ = 1/32 moles of O₂
We have already stated that 1/64 moles of oxygen contain x molecules. Therefore,
1/32 mole of oxygen will contain = x × 64 / 32 = 2x
Hence, 1 gm of oxygen consists of 2x number of molecules.

Answer:

Given:
Mass of one molecule of a substance = 4.65 × 10^-23
We know that the mass of a molecule of a substance is equal to the mass of 1 mole of that substance. 1 mole of a substance consists of 6.022 × 10²³ molecules; therefore, the mass of a molecule of a substance is equal to the mass of 6.022 × 10²³ molecules.
Therefore, to calculate the molecular mass of the substance, we need to calculate the mass of 6.022 × 10²³ molecules of that substance.
Mass of 1 molecule of the substance = 4.65 × 10^-23
Mass of 6.022 × 10²³ molecules of the substance = 6.022 × 10²³ × 4.65 × 10^-23 = 28g
Hence, molecular mass of the substance is 28 u.
Nitrogen has a molecular mass of 28 u; therefore, the substance is nitrogen.

Answer:

Given:
Mass of sulphur dioxide = 10g
Molar mass of sulphur dioxide = (1 × 32) + (2 × 16) = 64g
We know that 64g of sulphur dioxide contain 6.022 × 10²³ molecules.
Number of molecules in 64g  sulphur dioxide = 6.022 × 10²³
Number of molecules in 10g  sulphur dioxide  = 10 × 6.022 × 10²³/ 64
Hence, the number of molecules in 10g sulphur dioxide is 9.4× 10²².
Given:
Mass of oxygen = 10g
Molar mass of oxygen molecule = (2 × 16) = 32g
Number of molecules in 32g oxygen = 6.022 × 10²³
Number of molecules in 10g oxygen = 10 × 6.022 × 10²³/ 32
Hence, the number of molecules in 10g  oxygen is 18.8 × 10²²
It is clear that the number of atoms in 10g  oxygen is greater than that in 10g  sulphur dioxide.

Answer:

Mass of nitrogen gas = 56g
We know that equal moles of all substances contain equal number of molecules. (1 mole of all substances contains 6.022 × 10²³ molecules).
The first step will be to convert 56g  nitrogen gas into moles.
1 mole  nitrogen gas N₂ = 28g
28g  nitrogen = 1 mole
56g  nitrogen = 2 moles
Since, equal moles of all substances contain equal number of molecules, 2 moles of nitrogen will have the same number of molecules as 2 moles of oxygen.
The next step will be to find the mass of 2 moles of oxygen in grams.
1 mole of oxygen = 32g
2 moles of oxygen = 32 × 2 = 64g
Hence, 64g  oxygen will contain the same number of molecules as 56g  nitrogen gas.

Answer:

Mass of water = 1.8g
We know that equal moles of all substance contain equal number of molecules (1 mole of all substances contains 6.022 × 10²³ molecules).
The first step will be to convert 1.8g  water into moles.
1 mole of water = (2 × 1) + (1 × 16) = 18g
18g water = 1 mole
1.8g  water = 0.1 moles
Since, equal moles of all substances contain equal number of molecules, 0.1 moles of water will have the same number of molecules as present in 0.1 moles of nitrogen.
The next step will be to find the mass of 0.1 moles of nitrogen in grams.
1 mole of nitrogen = 28g
0.1 moles of nitrogen     = 28 × 0.1 = 2.8g
Hence, 2.8g  nitrogen will contain the same number of molecules as present in 1.8g  water.

Answer:

Given:
1g  sulphur contains x atoms
We know that equal moles of all substances contain equal number of molecules (1 mole of all substances contains 6.022 × 10²³ molecules).
Therefore, the ratio of molecules in sulphur and oxygen will be the same as their ratio of moles.
Therefore, 1g of S = 1/32 moles of S
Now, 1/32 moles of S contain x molecules (given). Since equal number of molecules are present in equal moles of all substances, 1/32 moles of oxygen will also contain x number of molecules.
Number of moles in 1g of oxygen:
1 mole of O = 16g
16g  O = 1 mole of O
Therefore, 1g  O₂ = 1/16 moles of O₂
We have already stated that 1/32 moles of oxygen contain x molecules; therefore,
1/16 mole of oxygen will contain = x × 32 / 16 = 2x molecules
Hence, 1 gram  oxygen consists of 2x the number of sulphur molecules.

Answer:

Mass of carbon = 6g
We know that equal moles of all substances contain equal number of atoms (1 mole of all substances contains 6.022 × 10²³ atoms).
The first step will be to convert 6g  carbon into moles.
1 mole of carbon = 12g
12g  carbon = 1 mole
Therefore, 6g carbon = 0.5 moles
Since, equal moles of all substances contain equal number of atoms, 0.5 moles of carbon will have the same number of atoms as present in 0.5 moles of magnesium.
The next step will be to find out the mass of 0.5 moles of magnesium in grams.
1 mole of magnesium = 24g
0.5 moles of magnesium = 24 × 0.5 = 12g
Hence, 12g magnesium will contain the same number of atoms as present in 6g  carbon.

Answer:

Given:
Mass of an atom of element X = 2.0 × 10^-23g
We know that the mass of an atom of a substance is equal to the mass of 1 mole of that substance. 1 mole of a substance consists of 6.022 × 10²³ atoms; therefore, the mass of an atom of a substance is equal to the mass of 6.022 × 10²³ atoms.
Therefore, to calculate the atomic mass of element X, we need to calculate the mass of 6.022 × 10²³ atoms of X.
Mass of 1 atom of X = 2.0 × 10^-23
Mass of 6.022 × 10²³ atoms of X = 6.022 × 10²³ × 2.0 × 10^-23 = 12 u
Hence, the atomic mass of element X is 12 u.
Carbon has an atomic mass of 12 u; therefore,  element X is carbon.